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Consider a matrix $\left[\begin{array}{lll}0 & 1 & 0 \\ a & 2 & d \\ b & 3 & c\end{array}\right]$. The matrix cannot have rank.

  1. $0$
  2. $1$
  3. $2$
  4. $3$

     

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  • Rank of a matrix: The rank of a matrix represents the maximum number of linearly independent rows or columns in the matrix.
  • Zero rank: A matrix has rank 0 only if all its elements are zero.

In the given matrix, even when a = b = c = 0, the second row still has a non-zero element (2). Therefore, it cannot have rank 0, as it has at least one linearly independent row.

 

  • Rank 1: If all rows are linearly dependent (e.g., a = 0, b = 1, c = 2), the matrix will have rank 1.
  • Rank 2: If there are two linearly independent rows (e.g., a = 1, b = 0, c = 1), the matrix will have rank 2.
  • Rank 3: If all three rows are linearly independent (e.g., a = 1, b = 2, c = 3), the matrix will have rank 3.
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using row operations to get

$$\begin{array} {|r|r|}\hline 0 & 1 & 0 \\ \hline a & 0 & d \\ \hline b & 0 & c \\ \hline  \end{array}$$

  1. now put a=b=c=d=0
    we get rank=1 $$\begin{array} {|r|r|}\hline 0 & 1 & 0 \\ \hline 0 & 0 & 0 \\ \hline 0 & 0 & 0 \\ \hline  \end{array}$$
  2. put a=d=c=0 and b $\ne$ 0
    we get rank=2 $$\begin{array} {|r|r|}\hline 0 & 1 & 0 \\ \hline 0 & 0 & 0 \\ \hline b & 0 & 0 \\ \hline  \end{array}$$
  3. put b = 0 and d =0  and a $\ne$ 0 and c $\ne$ 0
    we get rank=3 $$\begin{array} {|r|r|}\hline 0 & 1 & 0 \\ \hline a & 0 & 0 \\ \hline 0 & 0 & c \\ \hline  \end{array}$$
  4. proving rank is always atleast 1
    converting original matrix to row echelon form
    r2 = r2 – 2r1
    r3 = r3 – 3r1
    r3 = r3 – (b/a)r2
    r2 = r2 – $\frac{d}{c – (bd)/a}$r3
    switch r1 and r2


    $$\begin{array} {|r|r|}\hline a & 0 & 0 \\ \hline 0 & 1 & 0\\ \hline 0 & 0 & c - (bd)/a \\ \hline  \end{array}$$

    as you can see row 2 will always persist no matter what so rank is always at least 1

    so ANS – A
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Ans isA( 0) , at any value of a,b,c,d , the value of rank of matrix can't be zero .

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